15 Exponential Functions and Logarithms

15

Exponential Functions

and Logarithms

Objectives

To define and understandexponential functions.

Tosketch graphsof the various types of exponential functions. To understand the rules for manipulatingexponential and logarithmic expressions.

To solveexponential equations. To evaluatelogarithmic expressions.

To solve equations usinglogarithmic methods.

To sketch graphs of functions of the formy=logaxand simple transformations

of this.

To understand and use a range ofexponential models. Tosketch graphsof exponential functions.

Toapplyexponential functions to solving problems.

The function f(x)=kax_{, wherekis a non zero constant and the baseais a positive real}

number other than 1, is called anexponential function(or index function).

Consider the following example of an exponential function. Assume that a particular biological organism reproduces by dividing every minute. The following table shows the population,P, afternone-minute intervals (assuming that all organisms are still alive).

n 0 1 2 3 4 5 6 n

P 1 2 4 8 16 32 64 2n

ThusPdefines a function which has the ruleP=2n_{, an exponential (or index) function.}

15.1

Graphs of exponential functions

Two types of graphs will be examined.

1 Graphs of

y

=

a

,

a

> 1

Example 1

Plot the graph ofy=2x_{, and examine the table of values for−3≤x ≤3. A calculator can}

be used.

Solution

x −3 −2 −1 0 1 2 3

y=2x 0.125 0.25 0.5 1 2 4 8 8 6 4 2

–3 –2 –1 0 1 2 3

x y = 2x y

It can be observed that, as negative values of increasing magnitude are considered, the graph approaches thex-axis from above. Thex-axis is said to be anasymptote. As the magnitude of negativex-values becomes larger, 2x_{takes values closer and closer to}

zero, but never reaches zero, i.e. the graph gets closer and closer to thex-axis. This is written asx→ −∞,y→0 from the positive side, or asx→ −∞,y→0+.

They-axis intercept for the graph is (0, 1). The range of the function isR+.

Example 2

Plot the graph ofy=10x_{and examine the table of values for−1≤x ≤1. A calculator can}

be used.

Solution

x −1 −0.5 0 0.5 1

y=10x 0.1 0.316 1 3.16 10

y

10 8 6 4 2 1

–1 – 0.5 0 0.5 1

x y = 10x

Thex-axis is an asymptote, and they-axis intercept is (0, 1). They-values increase as thex-values increase. This rate of increase fory=10x_{is greater than that}

It is worth noting at this stage that, foraandbpositive numbers greater than 1, there is a positive numberksuch thatak _{=b. This can be seen from the graphs ofy=2}x _andy=10x

above. Using a calculator to solve 2k =10 graphically givesk=3.321928. . .Hence

10x₌₍₂3.321928 . . .₎x₌₂3.321928 . . .x_{. This means that the graph ofy=10}x_{can be obtained from}

the graph ofy=2xby a dilation of factor 1

k =

1

3.321928. . . from they-axis.

This shows that all graphs of the formy=ax_{, wherea>1, are related to each other by}

dilations from they-axis.

This will be discussed again later in the chapter.

2 Graphs of

y

=

a

, 0 <

a

< 1

Example 3

Plot the graph ofy=1₂xand examine the table of values for−3≤x≤3. A calculator can be used.

Solution

x −3 −2 −1 0 1 2 3

y=1₂x =2−x 8 4 2 1 0.5 0.25 0.125

Thex-axis is an asymptote, and they-axis intercept is (0, 1). For this graph the

y-values decrease as thex-values increase. This is written asx→ ∞,

y→0 from the positive side, or as

x→ ∞,y→0+. The range of the function isR+.

8 6 4 2

–3 –2 –1 0 1 2 3

y

x 2

1

y= = 2–x

x

In general:

y

x y= ax_{, a > 1}

1

0

y

x y= ax_{, 0 < a < 1}

1

0

The graph ofy=a−x _{is obtained from the graph ofy =a}x _{by a reflection in they-axis.}

Thus, for example, the graph ofy=1₂xis obtained from the graph ofy=2xby a reflection in they-axis, and vice versa.

Ifa=1, the resulting graph is a horizontal line with equationy=1.

Using a CAS calculator

Example 4

Plot the graph of y=2x_{on a CAS calculator and hence find:} a the value ofywhenx=2.1, correct to 3 decimal places

b the value ofxwheny=9.

Solution

Plot the graph for−3≤x ≤4.

a Select1:valuefrom theF5 (MATH) menu and enterx=2.1.

y=4.287 whenx=2.1 (Answer correct to 3 decimal places.)

b Enter y2=9 in theY=window. Return to theGRAPHmenu and select5:intersect from theF5 (MATH)menu.

Wheny=9,x=3.170, correct to 3 decimal places.

Transformations of exponential functions

Example 5

Sketch the graphs of each of the following functions. Give equations of asymptotes and the y-axis intercepts, and state the range of each of the functions. (x-axis intercepts need not be given.)

a f:R→R, f(x)=2x+3

b f:R→R, f(x)=2×3x₊₁ c f:R→R, f(x)= −3x+2

Solution

a For the function f:R→R, f(x)=2x_{+3, the corresponding graph is obtained}

by transforming the graph ofy=2x_{by a translation of 3 units in the positive}

The asymptote ofy=2x, with equationy=0, is transformed to the asymptote with equationy=3 for the graph of f(x)=2x_{+3. The asymptotic behaviour can}

be described asx →−∞,y→3 from the positive side, or asx →−∞,y→3+. As f(0)=20_{+3=4, they-axis intercept is 4.}

The range of the function f:R→R, f(x)=2x_{+3 is (3,∞).}

0 x

y

y= 2x

y= 2x _{+ 3}

y= 3

4

b For the function f:R→R, f(x)=2×3x+1, the corresponding graph is obtained by transforming the graph ofy=3x_{by a dilation of factor 2 from the} x-axis, followed by a translation of 1 unit in the positive direction of they-axis.

The asymptote ofy=3x_{, with equationy=0, is transformed to the asymptote} y=1 for the graph of f(x)=2×3x _+1.

The asymptotic behaviour is described asx →−∞,y→1 from the positive side, or asx →−∞,y→1+. They-axis intercept is given by

f(0)=2×30+1=3. The range of the function f:R→R, f(x)=2×3x+1 is (1,∞).

x y

0

y= 1

y= 3x

y= 2 × 3x_{+ 1}

3

c For the function f:R→R, f(x)= −3x_{+2, the corresponding graph is obtained}

by transforming the graph ofy=3xby reflection in thex-axis followed by a translation of 2 units in the positive direction of they-axis.

The asymptote ofy=3x, with equationy=0, is transformed to the asymptote

The asymptotic behaviour is described by asx →−∞,y→2 from the negative side, or asx→ −∞,y→2−. They-axis intercept is given by

f(0)= −30+2=1. The range of the function f:R→R, f(x)= −3x+2 is (−∞, 2).

x y

y = 2

y= 3x

y= –3x_{+ 2}

2 0

Example 6

Sketch the graphs of each of the following:

a y=2×3x _{b y=3}2x _{c y=3}x_{2 d y}= −₃2x₊₄

x y

y= 0

y= 3x

1

1 0

(1, 3) (1, 6)

2

Solution

a The graph of y=2×3xis obtained from the graph ofy=3x_{by a dilation of factor 2}

from thex-axis.

The horizontal asymptote for both graphs has equationy=0.

b The graph of y=32x _{is obtained from}

the graph ofy=3xby a dilation of factor 1₂ from they-axis. (In the notation introduced in

Chapter 6, write it as (x,y)→1₂x,y. Then describe the

transformation asx= 1₂xandy=y, and hencex=2xandy=y. The graph ofy=3xis mapped to the graph ofy=32x₎

The horizontal asymptote for both graphs has equationy=0.

x y

y = 0

1 1 0

(1, 3) , 3

y= 32x

y= 3x

c The graph ofy=3

x

2 _{is obtained}

from the graph ofy=3x_{by a}

dilation of factor 2 from they-axis. (In the notation introduced in

Chapter 6, write it as (x,y)→(2x,y).) Then describe the transformation as

x=2xandy=y, and hence

x= x

2 andy=y

_{. The graph ofy=3}x

is mapped to the graph ofy=3x

2).

x y

y= 0

1

0 (1, 3)

y = 3 y = 3x

(2, 3) x 2

d The graph ofy= −32x_{+4 is}

obtained from the graph ofy=3xby a dilation of factor1₂ from they-axis, followed by a reflection in thex-axis then by a translation of 4 units in the positive direction of they-axis.

x y

y= 4 y= 3x

y= –32x_{+ 4} 1

0 3

Exercise

15A

1 Using a calculator, plot the graphs of the following and comment on the similarities and differences between them:

a y=1.8x b y=2.4x _{c y}=₀.₉x _{d y=0.5}x

2 Using a calculator, plot the graphs of the following and comment on the similarities and differences between them:

a y=2×3x _{b y}=₅×₃x _{c y}= −₂×₃x _{d y= −5×3}x

3 Using a calculator plot the graph ofy=2x forx∈[−4, 4] and hence find the solution of

Example 4

the equation 2x_=14.

4 Using a calculator plot the graph ofy=10x_{forx∈[−0.4, 0.8] and hence find the solution}

of the equation 10x=6.

5 Sketch the graphs of each of the following functions. Give equations of asymptotes and

Example 5

y-axis intercepts, and state the range of each of the functions. (x-axis intercepts need not be given.)

a f:R→R, f(x)=3×2x+2 b f:R→R, f(x)=3×2x−3

c f:R→R, f(x)= −3x_{−2 d f:R→R, f(x)= −2×3}x₊₂

e f:R→R, f(x)=

1 2

x

+2 f f:R→R, f(x)= −2×3x₋₂

6 Sketch the graphs of each of the following:

Example 6

15.2

Reviewing rules for exponents (indices)

We shall review the rules for manipulating the exponential expressionax_{, whereais a real}

number called thebase, andxis a real number called theexponentorindex. Other words which are synonyms forindexarepowerandlogarithm.

Multiplication:

a

x

a

Ifmandnare positive integers

then am =a×a×a· · · ×a ←mterms→ and an =a×a· · · ×a

←nterms→

∴ am_×an _{=(a×a×a· · · ×a)×(a×a· · · ×a)} ←mterms→ ←nterms→

=(a×a×a· · · ×a)

←m+nterms→

=am+n

Rule 1

To multiply two numbers in exponent form with thesame base,addthe exponents.

am×an =am+n

Example 7

Simplify each of the following:

a 23×212 b x2_y3_×x4_{y c 2}x_×2x+2 _{d 3a}2_b3_×4a3_b2

Solution

a 23×212=23+12

=215

b x2y3×x4y=x2×x4×y3×y =x6_y4

c 2x×2x+2=2x+x+2

=22x+2

d 3a2b3×4a3b2=3×4a2×a3×b3×b2 =12a5_b5

Using a CAS calculator

c and d

Division:

a

÷

a

Ifmandnare positive integers andm>n ←mterms→

then am_÷an ₌ a×a×a. . .a a×a×a. . .a ←nterms→

=a×a×a. . .a (by cancelling) ←(m−n) terms→

=am−n

Rule 2

To divide two numbers in exponent form with thesame base,subtractthe exponents.

am÷an =am−n

Definea0 =1 fora=0 anda−n = 1

an fora=0.

Note:Rule 1andRule 2also hold for negative indicesm,nfora=0. For example:

24×2−2 = 2

4

22 =

2×2×2×2 2×2 =2

2

(i.e.24+(−2))

2−4÷22 = 1 24 ×

1 22 =

1 24_×22 =2

−6 _(i.e.2−4−2₎

23÷23 =23× 1

23 =1=2

0 _(i.e.23−3₎

Example 8

Simplify each of the following:

a x

4_y3

x2_y2 b

b4x×bx+1

b2x c

16a5_b×4a4_b3

8ab

Solution

a x 4_y3

x2_y2 =x

4−2_y3−2

=x2_y

b b

4x_×bx+1

b2x =b

4x+x+1−2x

=b3x+1

c 16a

5_b×4a4_b3

8ab =

16×4 8 ×a

5+4−1_×b1+3−1

Raising the power: (

a

)

Consider the following:

(23)2=23×23 =23+3=26=23×2

(43₎4₌₄3_×43_×43_×43₌₄3+3+3+3₌₄12₌₄3×4

(a2₎5 _=a2_×a2_×a2_×a2_×a2 _=a2+2+2+2+2_=a10_=a2×5

In general (am)n =am×n.

Rule 3

To raise the power ofato another power,multiplythe indices.

(am)n =am×n

This rule holds for all integersmandn.

Products and quotients

Consider (ab)n_anda b

n

.

(ab)n =(ab)×(ab)× · · · ×(ab)

← nterms →

=(a×a× · · ·a)×(b×b× · · ·b)

←nterms → ←nterms →

=anbn

Rule 4

(ab)n =anbn

_a

b

n = a

b × a b× · · ·

a b

= an bn

Rule 5

_a

b

n = an

bn

Example 9

Simplify the following, expressing the answers in positive exponent form:

a 8−2 b

1 2

−4

c 3

−3_×64_×12−3

9−4_×2−2 d

Solution

a 8−2 ₌ 1

82

= 1

(23₎2

= 1

26

b

1 2

−4

= 1

2−4

=24

c 3

−3_×64_×12−3

9−4×₂−2 =

3−3_×24_×34_×3−3_×2−3_×2−3

3−4×₃−4×₂−2

= 3−2×2−2

3−8_×2−2

=36

d 3 2n_×6n

8n×₃n =

(3n_×3n_)×(3n _×2n₎

23n×₃n =

3n_×3n

22n =

3 2

2n

Exercise

15B

1 Use the stated rule for each of the following to give an equivalent expression in simplest form:

a x2×x3 b 2×x3×x4×4 Rule 1

c x 5

x3 d

4x6

2x3 Rule 2

e (a3₎2 _{f (2}3₎2 _{Rule 3}

g (x y)2 h (x2y3)2 Rule 4 (also use Rule 3 for h)

i

x y

3

j

x3

y2

2

Rule 5 (also use Rule 3 for j)

2 Simplify the following:

Example 7 _{a x}3_×x4_×x2 _{b 2}4_×43_×82 _{c 3}4 _×92_×273

d (q2_p)3 _×(qp3₎2 _{e a}2_b−3_×(a3_b2₎3 _{f (2x}3₎2_×(4x4₎3 g m3p2×(m2n3)4×(p−2)2 h 23a3b2×(2a−1b2)−2

3 Simplify the following:

Example 8 _a x3y5

x y2 b

16a5b×4a4b3

8ab c (−2x y)

2_×2(x2_y)3

8(x y)3 d

(−3x2y3)2 (2x y)3 ×

4x4y3

4 Simplify each of the following, expressing your answer in positive exponent form:

Example9

a m3n2p−2×(mn2p)−3 b x

3_yz−2_×2(x3_y−2_z)2

x yz−1 c

a2b×(ab−2)−3 (a−2_b−1₎−2

d a

2_b3_c−4

a−1_b2_c−3 e

a2n−1_×b3_×c1−n an−3×_b2−n×_c2−2n

5 Simplify each of the following:

a 34n×92n ×273n b 2

n _×8n+1

32n c

3n−1×92n−3 62×₃n+2

d 2

2n_×92n−1

6n−1 e

252n_×5n−1

52n+1 f

6x−3_×4x

3x+1

g 6

2n_×93

27n×₈n×₁₆n h

3n−2_×9n+1

27n−1 i

8×25_×37

9×27×₈₁ 6 Simplify and evaluate:

a (8 3₎4

(212₎2 b

(125)3

(25)2 c

(81)4÷273 92

15.3

Rational exponents

a

1

n,_{wherenis a natural number, is defined to be then}th_{root ofa, and denoted} √n_a._Ifa≥₀

thena

1

n _{is defined for alln}∈_{N. Ifa}<_{0 then}√n_{ais only defined fornodd. (Remember that}

only real numbers are being considered.)

an1 = √n _a, _with

a1n

n =a

Using this notation for square roots:

√

a= √2a=a

1 2

Further, the expressionax_{can be defined for rational exponents, i.e. whenx =} m

n, wherem

andnare integers, by defining

amn =

an1

m

Example 10

Evaluate:

a 9−12 b 1652 c 64−

2 3

Solution

a 9−12 = 1

9

1 2

= √1

9 = 1

3 b 16

5 2 = 16 1 2 5

=√2

16

5

=45 ₌₁₀₂₄

c 64−23 = 1

6423

= 1

64

1 3

2 = ₃√1

64

2 = ₄12 =

Earlier we stated the rules:

am×an =am+n am÷an =am−n

(am)n =am×n

wheremandnare integers.

Note:These rules are applicable for all rational exponents.

a m q ×_a

n p =_a

m q+ n p a m q ÷_a

n p =_a

m q− n p a m q n p =a m q× n p Example 11 Simplify: a 3 1

4 ×√6×√42 1634

b (x−2y)

1 2 ×

x y−3

4

Solution

a 3 1

4 ×√6×√42 16

3 4

= 3

1

4 ×312 ×212 ×214

16 1 4 3 = 3 1 4 ×₃

1 2 ×₂

1 2 ×₂

1 4

23 =

3

3 4 ×₂

3 4 23 = 3 3 4

2124− 3 4

= 3

3 4

294

b (x−2y)12 ×

x y−3

4

= x−1y12 × x 4

y−12

= x3×y252

Exercise

15C

1 Evaluate each of the following:

Example10

a 12523 b 243

3

5 c 81−

1

2 d 6423

e 1 8 1 3

f 32−25 g 125−

2

3 _{h 32}

4 5

i 1000−

4

3 j 10 00034 _k 8134 _l

27 125

2 Simplify:

Example11

a √3a2_b÷√_ab3 _{b (a}−2_b)3_×

1

b−3

1 2

c 45

1 3

934 ×1532

d 232 ×4−14 ×16−34 _e

x3_y−2

3−3_y−3

−2

÷

3−3_x−2_y

x4_y−2

2 f

5

√ a2

3 2

×√3

a5

1 5

3 Simplify each of the following:

a (2x−1)√2x−1 b (x−1)2√x−1 c (x2 +1)√x2+₁ d (x−1)√3 x−1 e √ 1

x−1 +

√

x−1 f (5x2₊₁₎√3_5x2+₁

15.4

Solving exponential equations and inequations

Method 1

Express both sides of the equation as exponents to the same base and then equate the exponents since, ifax=aythenx=y.

Example 12

Find the value ofxfor which:

a 4x=256 b 3x−1_{=81 c 5}2x−4 ₌₂₅−x+2

Solution

a 4x =256 4x ₌₄4

∴x=4

b 3x– 1₌₈₁

3x– 1=34

x– 1=4

∴x =5

c 52x– 4=25–x+2

=(52)–x+2

=5–2x+4

∴2x– 4=–2x+4 4x =8

x=2

Example 13

Solve 9x_=12×3x_−27.

Solution

(3x)2 =12×3x– 27 Let y =3x

Then y2 =12y– 27

y2_{– 12y+27=0}

(y– 3)(y– 9)=0

Therefore y– 3=0 or y– 9=0

y =3 or y=9

Method 2

Using a CAS calculator

Use1:solvefrom theAlgebramenu.

Example 14

Solve 5x=10 correct to 2 decimal places.

Solution

We use numerical solve by pressing ≈

to obtain the answer.

x=1.43 (correct to 2 decimal places)

Solution of inequalities

The property thatax>ay⇔x>y, wherea∈(1,∞), andax>ay⇔x<ywhena∈(0, 1) is used.

Example 15

Solve forxin each of the following:

a 16x_{>2 b 2}−3x+1 _< 1

16

Solution

a 24x>21

⇔4x>1

⇔ x> 1

4

b 2−3x+1_<2−4

⇔ −3x+1<−4

⇔ −3x<−5

⇔ x> 5

3

Exercise

15D

1 Solve forxin each of the following:

Example12

a 3x_{=27 b 4}x_{=64 c 49}x₌₇

d 16x=8 e 125x=5 f 5x=625

g 16x=256 h 4−x = 1

64 i 5

−x ₌ 1

125

2 Solve fornin each of the following:

a 5n ×252n−1=125 b 32n−4 =1 c 32n−1= 1 81

d 3

n−2

91−n =1 e 3

3n_×9−2n+1_{=27 f 2}−3n_×42n−2₌₁₆ g 2n−6₌₈2−n _{h 9}3n+3 ₌₂₇n−2 _{i 4}n+1 ₌₈n−2

j 322n+1 ₌₈4n−1 _{k 25}n+1 _{=5×390 625} l 1254−n ₌₅6−2n _{m 4}2−n ₌ 1

2048

3 Solve the following exponential equations:

a 2x−1_×42x+1 _{=32 b 3}2x−1_×9x _{=243 c (27×3}x₎2 ₌₂₇x_×3

1 2

4 Solve forx:

Example13

a 4(22x₎₌₈₍₂x_{)−4 b 8(2}2x₎₋₁₀₍₂x₎₊₂₌₀

c 3×22x−18(2x)+24=0 d 9x−4(3x)+3=0

5 Use a calculator to solve each of the following, correct to 2 decimal places:

Example14

a 2x=5 b 4x=6 c 10x=18 d 10x=56

6 Solve forxin each of the following:

Example15

a 7x>49 b 8x>2 c 25x≤5 d 3x+1 <81

e 92x+1 _{<243 f 4}2x+1_{>64 g 3}2x−2 _≤81

15.5

Logarithms

Consider the statement 23_=8.

This may be written in an alternative form: log₂8=3

which is read as ‘the logarithm of 8 to the base 2 is equal to 3’.

In general, ifa∈R+\{1}andx∈Rthen the statementsax=nand logan=xare

equivalent.

Further examples:

32_{=9 is equivalent to log} 39=2

104=10 000 is equivalent to log1010 000=4

a0_{=1 is equivalent to log}

a1=0

Example 16

Without the aid of a calculator evaluate the following:

a log232 b log381

Solution

a Let log232=x

2x=32 2x₌₂5

Thereforex=5, giving log232=5.

b Let log381=x

3x=81 3x₌₃4

Thereforex=4, giving log381=4.

Laws of logarithms

The index rules are used to establish other rules for computations with logarithms.

1 Letax=manday=n, wherem,nandaare positive real numbers.

∴ mn=ax×ay =ax+y

logamn=x+yand, sincex=logamandy=logan, it follows that:

log_amn=log_am+log_an Rule 1

For example:

log₁₀200+log₁₀5=log₁₀(200×5)

=log₁₀1000

=3

2 m n =

ax ay =a

x−y

∴log_a

_m

n

=x−yand so:

log_a

_m

n

=log_am−log_an Rule 2

For example:

log₂32−log₂8=log₂

32 8

=log₂4

3 Ifm =1,log_a

1

n

=log_a1−log_an = −log_an

Therefore

log_a

1

n

= −log_an Rule 3

4 mp =(ax)p

=ax p

∴ log_a(mp)=x p

and so

log_a(mp)= plog_am Rule 4 For example:

3 log₃4=log₃(43)=log₃64

Example 17

Without using a calculator simplify the following:

2 log₁₀3+log₁₀16−2 log₁₀

6 5

Solution

2 log₁₀3+log₁₀16−2 log₁₀

6 5

=log₁₀32+log₁₀16−log₁₀

6 5

2

=log₁₀9+log₁₀16−log₁₀

36 25

=log₁₀

9×16×25 36

=log₁₀100

=2

Example 18

Solve each of the following equations forx:

a log5x=3 b log5(2x+1)=2

c log2(2x+1) – log2(x−1)=4 d log3(x−1)+log3 (x+1)=1

Solution

b log5(2x+1)=2⇔2x+1=52

∴ 2x+1=25 2x=24

x=12

c log2(2x+1)−log2(x−1)=4

Then log₂

2x+1

x−1

=4

⇔ 2x+1

x−1 =2

4

2x+1=16(x−1) Then 17=14x

Therefore 17 14 =x

d log3(x−1)+log3(x+1)=1

Therefore log3[(x−1)(x+1)]=1

which impliesx2−1=3 andx= ±2

But the expression is not defined forx= −2, thereforex=2.

Exercise

15E

1 Use the stated rule for each of the following to give an equivalent expression in simplest form:

a log₂10+log₂a b log₁₀5+log₁₀2 Rule 1

c log₂9−log₂4 d log₂10−log₂5 Rule 2

e log₅

1 6

f log₅

1 25

Rule 3

g log₂(a3) h log₂(83) Rule 4

2 Without using a calculator evaluate each of the following:

Example16

a log327 b log5625 _{c log}

2

1 128

d log₄

1 64

e logxx4 f log20.125 g log1010 000 h log100.000 001 i −3 log5125

j −4 log162 k 2 log39 l −4 log164

3 Without using a calculator simplify each of the following:

Example17

a 1

2log1016+2 log105 b log216+log28

c log2128+log345−log35 d log432−log927 e logbb3−logb

√

b f 2 logxa+logxa3

g xlog2 8+log281−x h

3

2 loga a−loga

4 Solve forx:

Example18

a log39=x b log3x=3

c log5x= −3 d log10x=log104+log102

e log₁₀2+log₁₀5+log₁₀x−log₁₀3=2 f log10x= 1₂log1036−2 log103

g logx64=2 h log5(2x−3)=3

i log3(x+2)−log32=1 j logx0.01= −2

5 Solve each of the following forx.

a logx

1 25

= −2 b log4(2x−1)=3

c log4(x+2)−log46=1 d log4(3x+4)+log416=5 e log3(x2−3x−1)=0 f log3(x2−3x+1)=0

6 If log10x=aand log10y=c, express log10

⎛

⎝100x3y−

1 2

y2

⎞

⎠in terms ofaandc.

7 Prove that log₁₀

ab2

c

+log₁₀

c2

ab

−log₁₀(bc)=0.

8 If log_a

11 3

+log_a

490 297

−2 log_a

7 9

=log_a(k), findk.

9 Solve each of the following equations forx:

a log10(x2−2x+8)=2 log10x b log10(5x)−log10(3−2x)=1 c 3 log10(x−1)=log108 d log10(20x)−log10(x−8)=2 e 2 log105+log10(x+1)=1+log10(2x+7)

f 1+2 log10(x+1)=log10(2x+1)+log10(5x+8)

15.6

Using logarithms to solve exponential

equations and inequations

In Section 15.4 two methods were shown for solving exponential equations. Ifa∈R+\{1}andx∈Rthen the statementsax_{=nand log}

a n=xare equivalent. This

defining property of logarithms may be used in the solution of exponential equations.

Method 3—Using logarithms

Example 19

Solution

Take either the log10or loge(since these are the only logarithmic functions available

on your calculator) of both sides of the equation: Then log₁₀2x =log₁₀11

i.e. xlog₁₀2=log₁₀11 Therefore x= log1011

log₁₀2

Use your calculator to evaluate both the log1011 and log102.

Then x= 1.041

0.301 (to 3 decimal places)

≈3.46 (to 2 decimal places)

Example 20

Solve 32x– 1=28.

Solution

log₁₀32x−1 =log₁₀28 (2x−1) log₁₀3=log₁₀28 2x−1= log1028 log₁₀3

= 1.4472

0.4771 3.0331

∴2x =4.0331

x =2.017 (to 3 decimal places)

Example 21

Solve{x: 0.7x≥0.3}.

Solution

Taking log10of both sides:

log₁₀0.7x ≥log₁₀0.3

xlog₁₀0.7≥log₁₀0.3

x ≤ log100.3

log₁₀0.7 Note change of direction of inequality as log100.7<0.

x ≤ −0.5229 −0.1549

Example 22

Sketch the graph of f(x)=2×10x_{−4, giving the equation of the asymptote and the axes}

intercepts.

Solution

f(0)=2×100−4=2−4= −2 Asx →−∞,y→ −4 from the positive side, or asx→−∞,y→ −4+.

The equation of the horizontal asymptote isy= −4.

For thex-axis intercept consider 2×10x_−4=0.

Hence 2×10x=4 and thus 10x=2. Thereforex=log102=0.3010

(correct to 4 decimal places)

y

x y = –4

0 log₁₀(2)

–2

Exercise

15F

1 Solve each of the following equations correct to 2 decimal places:

a 2x =7 b 2x=0.4 c 3x=14

d 4x_{=3 e 2}−x_{=6 f 0.3}x₌₂

g 5x =3x−2 h 8x =2005x+1 i 3x−1=10

Example20

j 0.2x+1 _=0.6

2 Solve forx. Give values correct to 2 decimal places if necessary:

Example21

a 2x_{>8 b 3}x_{<5 c 0.3}x_>4

d 3x−1 ≤7 e 0.4x≤0.3

3 For each of the following sketch the graph ofy=f(x), giving the equation of the asymptote

Example22

and the axes intercepts:

a f(x)=2x_{−4 b f(x)=2×3}x _{−6 c f(x)=3×10}x₋₅

d f(x)= −2×10x_{+4 e f(x)= −3×2}x_{+6 f f(x)=5×2}x₋₆

4 In the initial period of its life a particular species of tree grows in the manner described by the ruled=d010mt, wheredis the diameter of the tree, in cm,tyears after the beginning of

15.7

Graph of

y

=

log

a

x, where a > 1

A table of values fory=log10xis given below (the values are correct to 2 decimal places).

Use your calculator to check these values.

x 0.1 1 2 3 4 5

y=log₁₀x −1 0 0.30 0.48 0.60 0.70

0.8

–0.8 0.6

–0.6 0.4

–0.4 0.2 –0.2 0

1 2 3 4 5

–1.0

y

x y = log₁₀x

Domain = R+_{, Range = R}

y = log₁₀x y = 10x

y = x

0 1 1

y

x

The graph of y = log₁₀xis the reflection in the line y=xof the graph of y = 10x_.

Note that log101=0 as 100=1 and asx→0+,y→ −∞.

The inverse of a one-to-one function was introduced in Section 6.7. The function f−1_{: (0,∞)→R, f}−1_(x)=log

10x

is the inverse function of f:R→R, f(x)=10x

In general,y=logaxis the rule for the inverse of the function with ruley =ax, where a>0 andx>0.

Note that loga(ax)=xfor allxandalogax =x for positive values ofx, as they are inverse

functions.

Fora>1, the graphs of logarithm functions all have this same shape.

Properties ofy=ax,a>1 domain=R

range=R+ a0=1

x →−∞,y→0+

Properties ofy=logax,a>1

domain=R+

range=R

loga1=0

x→0+,y→ −∞ ₀

y=ax

y= log_ax

y=x

1 1

x y

Example 23

Find the inverse of:

a y=102x _{b y=log}

Solution

a y=102x

Interchangingxandygives

x=102y

Therefore 2y=log10x

and y= 1

2log10x

b y=log10(2x)

Interchangingxandygives

x=log10(2y)

Therfore 10x_=2y

and y= 1

2×10

x

Example 24

Find the inverse of each of the following:

a f(x)=2x+3 b f(x)=log2 (x−2) c f(x)=5×2x+3

Solution

a Lety=2x_{+3 b Lety=log}

2(x−2)

Interchangingxandygives Interchangingxandygives

x=2y+3 x=log2(y−2)

∴ x−3=2y _{∴ 2}x_=y−2

and y=log2 (x−3) and y =2x +2

∴ f−1_(x)=log

2 (x−3) ∴ f−1(x)=2x+2

domain of f−1=(3,∞) domain of f−1=R

c Lety=5×2x₊₃

Interchangingxandygives

x=5×2y₊₃

∴ x−3

5 =2

y

and y=log2

x−3 5

∴ f−1_(x)=log 2

x−3 5

domain of f−1_=(3,∞)

Transformations can be applied to the graphs of logarithm functions. This is shown in the following example.

Example 25

Sketch the graphs of each of the following. Give the maximal domain, the equation of the asymptote and the axes intercepts.

Solution

a f(x)=log2(x−3)

For the function to be defined

x−3>0, i.e.x>3

The maximal domain is (3,∞)

x-axis intercept: log2(x−3)=0

impliesx−3=20_{, i.e.x=4}

Asymptote:x→3+,y→ −∞

y

x x= 3

4 0

b f(x)=log2 (x+2)

For the function to be defined

x+2>0, i.e.x>−2

The maximal domain is (−2,∞)

f(0)=log2(2)=1

x-axis intercept: log2(x+2)=0

impliesx+2=20, i.e.x= −1 Asymptote:x→ −2+,y→ −∞

x= –2

y

x

0 1

–1

c f(x)=log23x

For the function to be defined 3x>0, i.e.x>0

The maximal domain is (0,∞) For thex-axis intercept log2(3x)=0

implies 3x=20_{, i.e.x=} 1 3

Asymptote:x→0+,y→ −∞

y

x

0

x= 0

1 3

Exercise

15G

1 Sketch the graph of each of the following and state the domain and range for each:

a y=log10(2x) b y=2 log10x c y=log₁₀

1 2x

d y=2 log10(3x) e y=–log10x f y=log10(−x)

2 Determine the inverse of each of the following:

Example23

a y=100.5x b y=3 log10x c y=103x d y=2 log10(3x)

3 Find the rule for the inverse function of each of the following:

Example24

a f(x)=3x_{+2 b f(x)=log}

2(x−3) c f(x)=4×3x+2

d f(x)=5x−2 e f(x)=log2(3x) f f(x)=log2

_x

3

4 Sketch the graphs of each of the following. Give the maximal domain, the equation of the

Example25

asymptote and the axes intercepts:

a f(x)=log2(x−4) b f(x)=log2 (x+3) c f(x)=log2(2x) d f(x)=log2(x+2) e f(x)=log2

_x

3

f f(x)=log2(−2x)

5 Use a calculator to solve each of the following equations correct to 2 decimal places:

a 2−x =x b log10(x)+x=0

6 Use a calculator to plot the graphs ofy=log10(x2) andy=2 log10xfor

x∈[−10, 10],x=0.

7 On the same set of axes plot the graph ofy=log₁₀(√x) andy= 1

2log10xfor

x∈(0, 10].

8 Use a calculator to plot the graphs ofy=log10(2x)+log10(3x) and

y=log10(6x2)

9 Findaandksuch that the graph ofy=a10kx_{passes through the points (2, 6) and}

(5, 20).

15.8

Exponential models and applications

Fitting data

Using a CAS calculator

It is known that the points (1, 6) and (5, 96) lie on a curve with equationy=a×bx_.

Define f(x)=a×bx.

Then usesolve(f(1)=6 and f(5)=96,{a,b})|a>0 andb>0. The result is as shown.

The equations can also be solved foraandb‘by hand’:

a×b1=6 and a×b5=96

Divide the second equation by the first to obtainb4 _{=16. Henceb=2. Substitute in}

the first equation to obtaina=3.

Example 26

Take a rectangular piece of paper approximately 30 cm×6 cm. Fold the paper in half, successively, until you have folded it five times. Tabulate the times folded,f, and the number of creases in the paper,C.

Solution

Times folded, f 0 1 2 3 4 5 Creases,C 0 1 3 7 15 31

The rule connectingCandfis

C=2f_{−1, f∈N∪{0}.}

C

f

30 25 20 15 10 5

0 1 2 3 4 5

Example 27

The table below shows the increase in weight of Somu, an orang-utan born at the Eastern Plains Zoo. Draw a graph to show Somu’s weight increase for the first six months.

Months,m 0 1 2 3 4 5 6

Weight,w kg 1.65 1.7 2.2 3.0 3.7 4.2 4.8

Solution

Plotting these values:

4

3

2

1

0 1 2 3 4 5 6 m (months)

Note: It is appropriate in this case to form a continuous line.

Graph of

w = 1.65(1.2)m

Graph showing Somu’s weight increase

We shall now plot, on the same set of axes, the graph of the exponential function

w=1.65(1.2)m_{, 0≤m≤6.}

Table of values

m 0 1 2 3 4 5 6

w 1.65 1.98 2.38 2.85 3.42 4.1 4.93

It can be seen from the graphs that the exponential modelw=1.65(1.2)m_{approximates to the}

actual weight gain and would be a useful model to predict weight gains for any future

orang-utan births at the zoo. This model describes a growth rate for the first 6 months of 20% per month.

This problem can also be attempted with a CAS calculator.

Using a CAS calculator

Use theStat/List Editorapplication. Enter the data inlist 1andlist 2as shown.

From theCalc, F4 , menu choose3:Regression and then8:ExpReg.

Complete as shown.

Example 28

There are approximately ten times as many red kangaroos as grey kangaroos in a certain area. If the population of grey kangaroos increases at a rate of 11% per annum while that of the red kangaroos decreases at 5% per annum, find how many years must elapse before the

proportions are reversed, assuming the same rates continue to apply.

Solution

LetP=population of grey kangaroos at the start.

∴ number of grey kangaroos afternyears=P(1.11)n_,

and number of red kangaroos afternyears=10P(0.95)n_.

When the proportions are reversed

P(1.11)n =10×[10P(0.95)n] (1.11)n =100(0.95)n

Taking log10of both sides

log₁₀(1.11)n =log₁₀100(0.95)n

nlog₁₀1.11=log₁₀100+nlog₁₀0.95

n×0.04532=2+n(−0.0223)

n= 2

0.0676

=29.6

i.e. the proportions of kangaroo populations will be reversed by the 30th year.

Exponential growth

The above two examples are examples of exponential change. In the following,Ais a variable that is subject to exponential change.

LetAbe the quantity at timet. Then A= A0a, where A0is a positive constant andais a

real number.

Ifa>1 the model representsgrowth. Ifa<1 the model representsdecay. Physical situations in which this is applicable include:

the growth of cells population growth

continuously compounded interest radioactive decay

cooling of materials.

If there arencompound periods in a year, the interest paid per period is 5

n%. Therefore

at the end of the year the amount of the investment,A, is

A=10 000

1+ 5 100n

n

=10 000

1+ 1 20n

n

Enter the function Y1=(1+1/(20X))ˆX. Look at the table of values with an increment of one. It is found that the value approaches 1.05127 (correct to 5 decimal places) fornlarge. Hence, for continuous compounding it can be written thatA=10 000×(1.05126 . . .)x_{, where} xis the number of years of the investment.

Exercise

15H

1 Find an exponential model of the formy=abxto fit the following data.

Example27

x 0 2 4 5 10

y 1.5 0.5 0.17 0.09 0.006

2 Find an exponential model of the formp=abt_{to fit the following data}

t 0 2 4 6 8

p 2.5 4.56 8.3 15.12 27.56

3 A sheet of paper 0.2 mm thick is cut in half and one piece is stacked on top of the other.

a If this process is repeated complete the following table: Cuts,n Sheets Total thickness,T(mm)

0 1 0.2

1 2 0.4

2 4 0.8

3 8 ...

..

. ... ... 10

b Write down a formula which shows the relationship betweenTandn.

c Draw a graph ofTagainstnforn≤10.

d What would be the total thickness,T, after 30 cuts?

4 The populations (in millions),pandq, of two neighbouring American states, P and Q, over a period of 50 years from 1950 are modelled by functionsp=1.2×20.08t_and q=1.7×20.04t, wheretis the number of years since 1950.

a Plot the graphs of both functions using a calculator.

Review

Chapter summary

Groups of exponential functions Examples:

1

y

x

y

x

1

y = ax, a > 1 y = ax_{, 0 < a < 1}

Tomultiplytwo numbers in exponent form with the same base,addthe exponents:

am_×an _=am+n

Todividetwo numbers in exponent form with the same base,subtractthe exponents:

am÷an =am−n

To raise the power ofato another power,multiplythe exponents: (am)n =am×n

Ifax_=ay_thenx=y.

For rational exponents: a

1

n = √n _a

and amn =

a1n

m

Fora∈R+\{1}andx∈R:ax=yis equivalent to logay=x

Laws of logarithms

1 loga(mn)=logam+logan 2 loga

_m

n

=log_am−log_an

3 log_a

1

n

= −log_an 4 log_a(mp_{)= plog} am

Exponential equations can be solved by taking logarithms of both sides. e.g. If 2x₌₁₁

then log102x=log1011 and hencexcan be solved.

Multiple-choice questions

1 8x3÷4x−3=

A 2 B 2x0 _{C 2x}6 _{D 2x}−1 _E 2

x9

2 The expression a

2_b

(2ab2₎3 ÷

ab

16a0 =

A 2

a2_b6 B

2a2

b6 C 2a

2_b6 _D 2

ab6 E

1 128ab5 3 The function f:R→R,f(x)=3×2x_{−1 has range}

Review

4 The functionf:R+→R, wheref(x)=log23x, has an inverse function f−1. The rule for

f−1is given by

A f−1 _(x)=2x _{B f}−1_(x)=3x _{C f}−1_(x)= 1

32

x

D f−1 _(x)=23x _{E f}−1_(x)=log 2

1 3x

5 If log10(x−2)−3 log102x=1−log10y, thenyis equal to

A 80x 3

x−2 B 1+

8x3

x−2 C

60x x−2

D 1+ 6x

x−2 E 1−

x−2 8x3

6 The solution of the equation 5×25x_{=10 isxequals}

A 1

2 B

1

5 C

1

5log210 D 1

2 log25 E 1 52

5

7 The equation of the asymptote ofy=3 log2(5x)+2 is

A x=0 B x=2 C x=3 D x=5 E y=2

8 Which of the following graphs could be the graph of the function f(x)=2ax +b, wherea

andbare positive?

A

O x

y _B

O y

x

C

O x

y

D

O y

x

E

O y

x

9 Which one of the following functions has a graph with a vertical asymptote with equation

x=b?

A y=log2(x−b) B y=

1

x+b C y=

1

x+b −b

D y=2x_{+b E y}=₂x−b

10 The expression 2mh (3mh2₎3 ÷

mh

81m2 is equal to

A 6

mh6 B

6m2

h6 C 6m

2_h6 _D 6

m2_h6 E

Review

Short-answer questions (technology-free)

1 Simplify each of the following, expressing your answer with positive index:

a a

6

a2 b

b8

b10 c

m3_n4

m5_n6 d

a3b2

(ab2₎4

e 6a 8

4a2 f

10a7

6a9 g

8(a3₎2

(2a)3 h

m−1n2

(mn−2₎3

i (p−1q−2)2 _j (2a−4)3

5a−1 k

6a−1

3a−2 l

a4+a8 a2 2 Use logarithms to solve each of the following equations:

a 2x _{=7 b 2}2x_{=7 c 10}x_{=2 d 10}x_=3.6

e 10x=110 f 10x=1010 g 25x=100 h 2x=0.1

3 Evaluate each of the following:

a log264 b log10107 c loga a2 d log41

e log327 f log2

1 4

g log100.001 h log₂16 4 Express each of the following as single logarithms:

a log102+log103 b log104+2 log103−log106 c 2 log10a−log10b d 2 log10a−3−log1025 e log10x+log10y−log10x f 2 log10a+3 log10b−log10c 5 Solve each of the following forx:

a 3x(3x– 27)=0 b (2x– 8)(2x– 1)=0

c 22x ₋₂x+1_{=0 d 2}2x_{– 12×2}x₊₃₂₌₀

6 Sketch the graph of:

a y=2.2x _{b y= −3.2}x _{c y=5.2}−x

d y=2−x+1 e y=2x −1 f y=2x+2

7 Solve the equation log10x+log102x−log10(x+1)=0

8 Given 3x=4y=12z, show thatz= x y x+y.

9 Evaluate 2 log212+3 log25−log215−log2150. 10 a Given that logp7+logpk=0, findk.

b Given that 4 logq3+2 logq2 – logq144=2, findq.

11 Solve:

Review

Extended-response questions

1 This problem is based on the so-called ‘Tower of Hanoi’ puzzle. Given a number of different sized discs, the problem is to move a pile of discs (wheren,

the number of discs, is≥1) to a second location (if starting at

Athen either toBorC) according to the following rules:

location A

location B

location C

Only one disc can be moved at a time.

A total of only three locations can be used to ‘rest’ discs. A larger sized disc cannot be placed on top

of a smaller disc.

The task must be completed in the smallest possible number of moves.

a Using two coins complete the puzzle. Repeat first with three coins and then four coins and thus complete the table.

Number of discs,n 1 2 3 4 Minimum number of moves,M 1

b Work out the formula which shows the relationship betweenMandn. Use your formula to extend the table of values forn=5, 6 and 7.

c Plot the graph ofMagainstn.

d Investigate, for bothn=3 and 4, to find whether there is a pattern for the number of times each particular disc is moved.

2 To control an advanced electronic machine, 2187 different switch positions are required. There are two kinds of switches available:

Switch 1: These can be set in 9 different positions. Switch 2: These can be set to 3 different positions.

Ifnof switch type 1 andn+1 of switch type 2 are used, calculate the value ofnto give the required number of switch positions.

3 Research is being carried out to investigate the durability of paints of different thicknesses. The automatic machine shown in the diagram is proposed for producing a coat of paint of a particular thickness.

Blade set to reduce thickness

Thin layer Thick layer

Review

a Operating at the initial setting the blade reduces the paint thickness to one-eighth of the original thickness. This happensntimes. What fraction of the paint thickness remains? Express this as a power of 1₂.

b The blade is then reset so that it removes three-quarters of the remaining paint. This happens (n– 1) times. At the end of this second stage express the remaining thickness as a power of 1₂.

c The third phase of the process involves the blade being reset to remove half of the remaining paint. This happens (n– 3) times. At what value ofnwould the machine have to be set to reduce a film of paint 8192 units thick to 1 unit thick?

4 A hermit has little opportunity to replenish supplies of tea and so, to eke out supplies for as long as possible, he dries out the tea leaves after use and then stores the dried tea in an airtight box. He estimates that after each re-use of the leaves the amount of tannin in the used tea will be half the previous amount. He also estimates that the amount of caffeine in the used tea will be one-quarter of the previous amount.

The information on the label of the tea packet states that the tea contains 729 mg of caffeine and 128 mg of tannin.

a Write down expressions for the amount of caffeine when the tea leaves are